excess presuure of soap bubble of radius 1 cm is balanced by a column of oil of specific gravity 0.8 , 2mm high , the surface tension will be ? i got the answer but decimal is coming in the wrong place so its ok if u show just the calculation part n thanks in advance :)
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pressure at the depth of 2 mm in oil
= 0.8 * 1000 kg/m² * 9.81 m/sec² * 0.002 m
= 15.696 N/m²
P = P2 - P1
= excess pressure of a soap bubble = pressure inside - pressure outside
= 4 S / R , where S = surface tension and R = radius of the bubble.
S = P * R / 4 = 15.696 * 0.01 m / 4 = 0.03924 N/m
if there is some different answer, perhaps the given values of radii are not correct..
= 0.8 * 1000 kg/m² * 9.81 m/sec² * 0.002 m
= 15.696 N/m²
P = P2 - P1
= excess pressure of a soap bubble = pressure inside - pressure outside
= 4 S / R , where S = surface tension and R = radius of the bubble.
S = P * R / 4 = 15.696 * 0.01 m / 4 = 0.03924 N/m
if there is some different answer, perhaps the given values of radii are not correct..
kvnmurty:
click on thanks azur blue button above please
what answer did u get ?
0.3294 :(
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